Does MAPLE not simplify correctly or am I goofing?

Question

While attempting to answer this question with help of MAPLE, something very strange happened, at least according to me. We have the following function: $$ \phi(x,y) = x^2-y^2+\frac{x^2-y^2}{(x^2+y^2)^2} $$ The partial derivatives of this function are (according to MAPLE and myself): $$ u = \frac{\partial \phi}{\partial x} = 2 x + \frac{2 x}{(x^2+y^2)^2} - \frac{4(x^2-y^2)x}{(x^2+y^2)^3} \\ v = \frac{\partial \phi}{\partial y} = - 2 y - \frac{2 y}{(x^2+y^2)^2} - \frac{4(x^2-y^2)y}{(x^2+y^2)^3} $$ Then we calculate $\sqrt{u^2+v^2}$ and ask MAPLE to simplify, with the following result: $$ \sqrt{u^2+v^2} = 2\,\sqrt {{\frac{x^8+4\,x^6y^2-2\,x^4+6\,x^4y^4 +12\,x^2y^2+4\,x^2y^6-2\,y^4+y^8 \color{red}{+1}}{\left(x^2+y^2\right)^{3}}}} $$ But, even at first sight .. how can there be a $+1$ in that fraction ? Here comes the MAPLE sequence:

phi(x,y) := (x^2-y^2)+(x^2-y^2)/(x^2+y^2)^2;
diff(phi(x,y),x); diff(phi(x,y),y);
simplify(sqrt(diff(phi(x,y),x)^2+diff(phi(x,y),y)^2));

I use a legal MAPLE 8 version, to be precise.

Answer 1

Nothing wrong here, for example if you set $y=0$ and simplify by hand, you get

$$\sqrt{u^2} = 2\sqrt{\frac{(x^4-1)^2}{x^6}}$$

exactly what Maple says. You can see how $1$ emerges...

Answer 2

Maple 18 gives the same answer, which, as far as I can tell, is correct. For example, at $x=3, y=7$ both $\sqrt{u^2+v^2}$ calculated directly and Maple's answer simplify to $15\sqrt{50297} \sqrt{58}/1682$.

Answer 3

Here is a link for Maple bugs:

http://www.math.ncsu.edu/MapleInfo/MapleArchive/BugList.html

This is for maple V, but it was only a quick search. The moral of the story is be careful, and don't be so quick to discount drawnonward. I have had problems with Maple before....years ago. And I have a great memory. Oh, and I think all those down votes we certainly a bit hasty, as was my original answer. I apologize.